Changing basis of linear map The matrix wrt. the standard basis of a given linear map is:
$$[T]_B=\begin{bmatrix}-1&2&1\\3 & 1 & 0\\1&1&1\end{bmatrix}$$
I want to express this in the basis $B'={(0,1,2),(3,1,0),(0,1,1)}$.
Say I want to first find column one. Now, 
$$T(0,1,2)=T(0,1,0)+2 T(0,0,1)=\begin{bmatrix}4\\1\\3
\end{bmatrix}$$
To represent this in basis $B'$, i need to find the coefficients $\lambda_i$, such that:
$$\begin{bmatrix}4\\1\\3\end{bmatrix}=\lambda_1\begin{bmatrix}0\\1\\2\end{bmatrix}+\lambda_2\begin{bmatrix}3\\1\\0\end{bmatrix}+\lambda_{3}\begin{bmatrix}0\\1\\1\end{bmatrix}$$
First of all, I would like to hear if this is the right procedure to find the transformation matrix in basis $B'$. If yes, what is the easiest way to find the coefficients (yep, I'm new at this).
Edit:
I realized that my lecture notes states that if $\{x_i\}$ are the basis vectors of one basis and $\{y_i\}$ of another, then the transformation $Ax_k$ can be written in basis $\{y_i\}$ through: 
$$A x_{k}=\sum_{i} \alpha_{i k} y_{i}$$
In this case, $$Ax_1=[T]_Be_1=\begin{bmatrix}-1\\3\\1\end{bmatrix}=\alpha_{11} b_1+\alpha_{21} b_2+\alpha_{31} b_2=\alpha_{11}\begin{bmatrix}0\\1\\2
\end{bmatrix}+\alpha_{21}\begin{bmatrix}3\\1\\0\end{bmatrix}+\alpha_{31}\begin{bmatrix}0\\1\\1\end{bmatrix}$$
Which leads to $$\alpha_{11}=-\frac{7}{3}, \quad \alpha_{21}=-\frac{1}{3}, \quad \alpha_{31}=\frac{17}{3}$$
Is this correct, or did I misunderstand something?
 A: Yes, your thoughts are right. Note that
$$\left[\begin{array}{c}
4 \\ 1 \\ 3
\end{array}\right]=\lambda_{1}\left[\begin{array}{c}
0 \\ 1 \\ 2
\end{array}\right]+\lambda_{2}\left[\begin{array}{c}
3 \\ 1 \\ 0
\end{array}\right]+\lambda_{3}\left[\begin{array}{c}
0 \\ 1 \\ 1
\end{array}\right]$$
is equivalent to
$$\left[\begin{array}{c}
4 \\ 1 \\ 3
\end{array}\right]=\left[\begin{array}{c}
3\lambda_2 \\
\lambda_1+\lambda_2+\lambda_3 \\
2\lambda_1+\lambda_3
\end{array}\right]$$
which gives you the linear system
$$\left\{\begin{array}{rcrcrcc}
& & 3\lambda_2 & & & = & 4 \\
\lambda_1 & + & \lambda_2 & + & \lambda_3 & = & 1 \\
2\lambda_1 & & & + & \lambda_3 & = & 3
\end{array}\right.$$
Now you just need to solve it to find the first column of $[T]_{B'}$.
A: We have $[T]_{B'}^{B'} = [I_{\mathbb{R}^3}]_{B}^{B'}[T]_{B}^{B}[I_{\mathbb{R}^3}]_{B'}^{B}$ and 
$$Q:=[I_{\mathbb{R}^3}]_{B'}^{B} = \begin{bmatrix}
0 & 3 & 0\\
1 & 1 & 1\\
2 & 0 & 1
\end{bmatrix}.$$
Therefore, $[T]_{B'}^{B'} = Q^{-1}[T]_{B}^{B}Q$.
